Identify the errors in the proposed proof. Look at Selections in Description below and select all that apply. What is the correct answer?

Question

Consider the following statement. 3√(2) − 7 is irrational. This statement is true, but the following proposed proof by contradiction is incorrect.

Proposed proof:

Suppose not. That is, suppose 3√(2) − 7 is rational.

By definition of rational, there exist real numbers a and b with 3√(2) − 7 = a/b and b ≠ 0.

Using algebra to solve for √(2) gives 3b√(2) − 7b = a; and so √(2) = (a + 7b)/3b.

But a + 7b and 3b are real numbers and 3b ≠ 0.

Therefore, by definition of rational, √(2) is rational.

This contradicts Theorem 4.7.1, which states that √(2) is irrational. Hence the supposition is false.

Selections:

A. There is an error in the algebra solving for √(2).

B. For a proof by contradiction you should suppose that 3√(2) − 7 is irrational.

C. The square root of 2 is rational, so there is no contradiction.

D. The statement to be proved is assumed.

E. To apply the definition of rational, a and b must be integers.

Answer 1

The correct answer is E, and I can show you why this is important:

If we define rational numbers as used in the proof above, then a and b can be any real numbers (b =/= 0). We know that pi and 1 are both real numbers, and therefore pi / 1 = pi is rational. However, we know that pi is irrational, so does dividing by 1 make it rational? Of course not, and that's why this definition of rational numbers needs strengthening.